Given: - B = 3100 Hz

$\big(\frac{S}{N}\big)dB=10$

$∴10=10 log_{10}big(\frac{S}{N}\big) \\
∴\frac{S}{N}=10$

∴ Maximum bit rate

$= R_{\max} = B \log_2 \big[1+\frac{S}{N}\big] \\
= 3100 \log_2(1+10) \\
=\frac{3100 \log_{10}11}{\log_{10}2} \\
= 10,724 bits/sec$